Q21.17P - Chemistry: Molecular Nature Of Matter And Change

Question

Balance the following skeleton reactions and identify the oxidizing and reducing agents:

(a) $\mathit{N}{\mathit{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }\mathit{N}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\left(aq\right)\mathbf{+}\mathit{N}{\mathit{O}}_{\mathbf{2}}^{\mathbf{-}}\left(aq\right)\left[basic\right]$

(b) $\mathit{Z}\mathit{n}\left(s\right)\mathbf{ }\mathbf{+}\mathit{N}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}\left(aq\right)\mathbf{\to }\mathit{Z}\mathit{n}{{\left(OH\right)}_{\mathbf{4}}}^{\mathbf{2}\mathbf{-}}\left(aq\right)\mathbf{ }\mathbf{+}\mathit{N}{\mathit{H}}_{\mathbf{3}}\left(g\right)\left[basic\right]$

(c) ${\mathit{H}}_{\mathbf{2}}\mathit{S}\left(g\right)\mathbf{+}\mathbf{ }\mathit{N}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}\left(aq\right)\mathbf{\to }{\mathit{S}}_{\mathbf{8}}\left(s\right)\mathbf{+}\mathbf{ }\mathit{N}\mathit{O}\left(g\right)\left[acidic\right]$

Step-by-Step Solution

Answer

$$\begin{gathered} \left(a\right)2{\text{NO}}_2( \text{g})+2{\text{OH}}^{-}\to {\text{NO}}_3^{-}\left(\text{aq}\right)+{\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{H}}_2\text{O} \\ \left(b\right)4\text{Zn}\left(\text{s}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}_2\text{O}+7{\text{OH}}^{-}\to 4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+{\text{NH}}_3( \text{g}\left) \\ \right(c)24{\text{H}}_2 \text{S}( \text{g})+16{\text{NO}}_3^{-}(aq)+16{\text{H}}^{+}\to 3 {\text{S}}_8( \text{s})+16\text{NO}(\text{g})+32{\text{H}}_2\text{O} \end{gathered}$$

Definition of Oxidising Agent

In a redox chemical reaction, an oxidizing agent obtains or "accepts"/"receives" an electron from a reducing agent. In other words, any substance that oxidizes another substance is an oxidant.

Balancing of the given reaction in part a.

(a) ${\text{NO}}_2( \text{g})\to {\text{NO}}_3^{-}\left(\text{aq}\right)+{\text{NO}}_2^{-}\left(\text{aq}\right)\left[\text{basic}\right]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

 $$\begin{gathered} {\text{NO}}_2\left(g\right)\to {\text{NO}}_2^{-}\left(aq\right) \\ {\text{NO}}_2( \text{g})\to {\text{NO}}_3^{-}\left(aq\right) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{NO}}_2\left(g\right)\to {\text{NO}}_2^{-}\left(aq\right) \\ {\text{NO}}_2\left(g\right)\to {\text{NO}}_3^{-}\left(aq\right) \end{gathered}$$

3: Now add $H_{2}O$ molecules to balance oxygen atoms.

$$\begin{gathered} {\text{NO}}_2\left(g\right)\to {\text{NO}}_2^{-}\left(aq\right) \\ {\text{NO}}_2\left(g\right)+{\text{H}}_2\text{O}\to {\text{NO}}_3^{-}\left(aq\right) \end{gathered}$$

Balancing of the given reaction in part a.

4: To balance hydrogen atoms we add protons, $H^{+}$..

$$\begin{gathered} {\text{NO}}_2( \text{g})\to {\text{NO}}_2^{-}\left(aq\right) \\ {\text{NO}}_2( \text{g})+{\text{H}}_2\text{O}\to {\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{+} \end{gathered}$$

5: Now balance the charge with electrons, $e^{-}$..

$$\begin{gathered} {\text{NO}}_2( \text{g})+{\text{e}}^{-}\to {\text{NO}}_2^{-}\left(\text{aq}\right) \\ {\text{NO}}_2( \text{g})+{\text{H}}_2\text{O}\to {\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{+}+{\text{e}}^{-} \end{gathered}$$

6: Scale the reactions to make the electron count equal.

$$\begin{gathered} {\text{NO}}_2( \text{g})+{\text{e}}^{-}\to {\text{NO}}_2^{-}\left(aq\right) \\ {\text{NO}}_2( \text{g})+{\text{H}}_2\text{O}\to {\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{+}+{\text{e}}^{-} \end{gathered}$$

7: Add the reactions.

${\text{NO}}_2( \text{g})+{\text{e}}^{-}+{\text{NO}}_2( \text{g})+{\text{H}}_2\text{O}\to {\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{+}+{\text{e}}^{-}$

Balancing of the given reaction in part a.

8: Cancel out common terms.

$2{\text{NO}}_2( \text{g})+{\text{H}}_2\text{O}\to {\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{+}$

9: Since, the reaction is in basic medium we add $O{H}^{-}$ to balance the $H^{+}$ ions.

$2{\text{NO}}_2( \text{g})+{\text{H}}_2\text{O}+2{\text{OH}}^{-}\to {\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}^{+}+2{\text{OH}}^{-}$

10: Combine $O{H}^{-}$ ions and $H^{+}$ ions present on the same side to form water molecule.

$2{\text{NO}}_2( \text{g})+{\text{H}}_2\text{O}+2{\text{OH}}^{-}\to {\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+2{\text{H}}_2\text{O}$

11: Cancel out common terms.

$2{\text{NO}}_2( \text{g})+2{\text{OH}}^{-}\to {\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+{\text{H}}_2\text{O}$

Hence, we get the following balanced reaction:

$2{\text{NO}}_2( \text{g})+2{\text{OH}}^{-}\to {\text{NO}}_3^{-}\left(\text{aq}\right)+{\text{NO}}_2^{-}\left(\text{aq}\right)+{\text{H}}_2\text{O}$

Oxidizing agent: ${\text{NO}}_2( \text{g})$

Reducing agent: ${\text{NO}}_2( \text{g})$

Balancing of the given reaction in part b.

(b) $\text{Zn}\left(\text{s}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)\to \text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+{\text{NH}}_3( \text{g}\left)\right[\text{basic}]$

To balance the equation, we will follow the following steps:

1: Separate the half-reactions.

$$\begin{gathered} {\text{NO}}_3^{-}\left(aq\right)\to {\text{NH}}_3( \text{g}) \\ \text{Zn}\left(\text{s}\right)\to \text{Zn}\left(\text{OH}{)}_4^{2-}\right(aq) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{NO}}_3^{-}\left(aq\right)\to {\text{NH}}_3\left(g\right) \\ \text{Zn}\left(\text{s}\right)\to \text{Zn}\left(\text{OH}{)}_4^{2-}\right(aq) \end{gathered}$$

3: Now add $H_{2}O$ molecules to balance oxygen atoms.

$$\begin{gathered} {\text{NO}}_3^{-}\left(aq\right)\to {\text{NH}}_3( \text{g})+3{\text{H}}_2\text{O} \\ \text{Zn}\left(\text{s}\right)+4{\text{H}}_2\text{O}\to \text{Zn}\left(\text{OH}{)}_4^{2-}\right(aq) \end{gathered}$$

4: To balance hydrogen atoms we add protons, $H^{+}$.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)+9{\text{H}}^{+}\to {\text{NH}}_3( \text{g})+3{\text{H}}_2\text{O} \\ \text{Zn}\left(\text{s}\right)+4{\text{H}}_2\text{O}\to \text{Zn}\left(\text{OH}{)}_4^{2-}\right(aq)+4{\text{H}}^{+} \end{gathered}$$

Balancing of the given reaction in part b.

5: Now balance the charge with electrons, $e^{-}$.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)+9{\text{H}}^{+}+8{\text{e}}^{-}\to {\text{NH}}_3( \text{g})+3{\text{H}}_2\text{O} \\ \text{Zn}\left(\text{s}\right)+4{\text{H}}_2\text{O}\to \text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+4{\text{H}}^{+}+2{\text{e}}^{-} \end{gathered}$$

6: Scale the reactions to make the electron count equal.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)+9{\text{H}}^{+}+8{\text{e}}^{-}\to {\text{NH}}_3( \text{g})+3{\text{H}}_2\text{O} \\ 4\text{Zn}\left(\text{s}\right)+16H_2\text{O}\to 4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(aq)+16H^{+}+8e^{-} \end{gathered}$$

7: Add the reactions.

${\text{NO}}_3^{-}\left(\text{aq}\right)+9{\text{H}}^{+}+8{\text{e}}^{-}+4\text{Zn}\left(\text{s}\right)+16{\text{H}}_2\text{O}\to {\text{NH}}_3( \text{g})+3{\text{H}}_2\text{O}+4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+16{\text{H}}^{+}+8{\text{e}}^{-}$

8: Cancel out common terms.

${\text{NO}}_3^{-}\left(\text{aq}\right)+4\text{Zn}\left(\text{s}\right)+13{\text{H}}_2\text{O}\to {\text{NH}}_3( \text{g})+4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+7{\text{H}}^{+}$

9: Since, the reaction is in basic medium we add $O{H}^{-}$ to balance the $H^{+}$ ions.

${\text{NO}}_3^{-}\left(\text{aq}\right)+4\text{Zn}\left(\text{s}\right)+13{\text{H}}_2\text{O}+7{\text{OH}}^{-}\to {\text{NH}}_3( \text{g})+4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+7{\text{H}}^{+}+7{\text{OH}}^{-}$

Balancing of the given reaction in part b.

10: Combine $O{H}^{-}$ ions and $H^{+}$ ions present on the same side to form water molecule.

${\text{NO}}_3^{-}\left(\text{aq}\right)+4\text{Zn}\left(\text{s}\right)+13{\text{H}}_2\text{O}+7{\text{OH}}^{-}\to {\text{NH}}_3( \text{g})+4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+7{\text{H}}_2\text{O}$

11: Cancel out common terms.

${\text{NO}}_3^{-}\left(\text{aq}\right)+4\text{Zn}\left(\text{s}\right)+6{\text{H}}_2\text{O}+7{\text{OH}}^{-}\to {\text{NH}}_3( \text{g})+4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})$

Hence, we get the following balanced reaction:

$4\text{Zn}\left(\text{s}\right)+{\text{NO}}_3^{-}\left(\text{aq}\right)+6{\text{H}}_2\text{O}+7{\text{OH}}^{-}\to 4\text{Zn}\left(\text{OH}{)}_4^{2-}\right(\text{aq})+{\text{NH}}_3( \text{g})$

Oxidizing agent: ${\text{NO}}_3^{-}\left(aq\right)$

Reducing agent: Zn(s)

Balancing of the given reaction in part c.

(c) ${\text{H}}_2 \text{S}( \text{g})+{\text{NO}}_3^{-}\left(\text{aq}\right)\to {\text{S}}_8( \text{s})+\text{NO}\left(\text{g}\right)\left[acidic\right]$

To balance the equation, we will follow following steps:

1: Separate the half-reactions.

$$\begin{gathered} {\text{NO}}_3^{-}\left(aq\right)\to \text{NO}\left(g\right) \\ {\text{H}}_2 \text{S}( \text{g})\to {\text{S}}_8( \text{s}) \end{gathered}$$

2: Balance all the elements other than Oxygen and Hydrogen.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)\to \text{NO}\left(\text{g}\right) \\ 8{\text{H}}_2 \text{S}( \text{g})\to {\text{S}}_8( \text{s}) \end{gathered}$$

3: Now add $H_{2}O$ molecules to balance oxygen atoms.

$$\begin{gathered} {\text{NO}}_3^{-}\left(aq\right)\to \text{NO}\left(\text{g}\right)+2{\text{H}}_{2}O \\ 8{\text{H}}_2 \text{S}( \text{g})\to {\text{S}}_8( \text{s}) \end{gathered}$$

4: To balance hydrogen atoms we add protons, $H^{+}$.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{+}\to \text{NO}\left(\text{g}\right)+2{\text{H}}_2\text{O} \\ 8{\text{H}}_2 \text{S}( \text{g})\to {\text{S}}_8( \text{s})+16{\text{H}}^{+} \end{gathered}$$

Balancing of the given reaction in part c.

5: Now balance the charge with electrons, $e^{-}$.

$$\begin{gathered} {\text{NO}}_3^{-}\left(\text{aq}\right)+4{\text{H}}^{+}+3{\text{e}}^{-}\to \text{NO}\left(\text{g}\right)+2{\text{H}}_2\text{O} \\ 8{\text{H}}_2 \text{S}( \text{g})\to {\text{S}}_8( \text{s})+16{\text{H}}^{+}+16{\text{e}}^{-} \end{gathered}$$

6: Scale the reactions to make the electron count equal.

$$\begin{gathered} 16{\text{NO}}_3^{-}\left(aq\right)+64H^{+}+48e^{-}\to 16\text{NO}\left(\text{g}\right)+32H_{2}O \\ 24H_2\text{S}( \text{g})\to 3{\text{S}}_8( \text{s})+48H^{+}+48e^{-} \end{gathered}$$

7: Add the reactions.

$16{\text{NO}}_3^{-}\left(\text{aq}\right)+64{\text{H}}^{+}+48{\text{e}}^{-}+24{\text{H}}_2 \text{S}( \text{g})\to 16\text{NO}\left(\text{g}\right)+32{\text{H}}_2\text{O}+3 {\text{S}}_8( \text{s})+48{\text{H}}^{+}+48{\text{e}}^{-}$

                                                                           

 

Balancing of the given reaction in part c.

8: Cancel out common terms.$16{\text{NO}}_3^{-}\left(\text{aq}\right)+16{\text{H}}^{+}+24{\text{H}}_2 \text{S}( \text{g})\to 16\text{NO}\left(\text{g}\right)+32{\text{H}}_2\text{O}+3 {\text{S}}_8( \text{s})$

Hence, we get the following balanced reaction:

$24{\text{H}}_2 \text{S}( \text{g})+16{\text{NO}}_3^{-}\left(\text{aq}\right)+16{\text{H}}^{+}\to 3 {\text{S}}_8( \text{s})+16\text{NO}\left(\text{g}\right)+32{\text{H}}_2\text{O}$

Oxidizing agent: ${\text{NO}}_3^{-}\left(aq\right)$

Reducing agent: ${\text{H}}_2 \text{S}( \text{g})$